D $3,920.00

Q4

Pipe A can fill a water tank in 6 hours, while Pipe B can fill the same tank in 12 hours. If both pipes are opened at the same time, how many hours will it take to fill the tank completely?

A 3 hours

B 4 hours

C 5 hours

D 9 hours

Texas Success Initiative Assessment 2.0 • Mathematics Practice Exam Page 2 of 6

Section 2: Algebraic Reasoning Questions 5 - 8

Q5

Which of the following is equivalent to the expression \(\frac{2x^2 - 8}{x^2 + 2x - 8}\) for all values of \(x\) where the expression is defined?

A \(\frac{2(x-2)}{x+4}\)

B \(\frac{2(x+2)}{x+4}\)

C \(\frac{2x}{x-4}\)

D \(\frac{x+2}{x+4}\)

Q6

Solve the inequality: \(-3(2x - 4) + 5 \ge 4x - 7\). Which of the following represents the correct solution set?

A \(x \le 2.4\)

B \(x \ge 2.4\)

C \(x \le 1.0\)

D \(x \ge 1.0\)

Q7

A line passes through the points \((-2, 5)\) and \((4, -13)\). What is the equation of this line in slope-intercept form?

A \(y = -3x - 1\)

B \(y = -3x + 1\)

C \(y = 3x - 1\)

D \(y = -\frac{1}{3}x + 3\)

Q8

Solve the system of equations: \(3x - 4y = 18\) and \(2x + 3y = -5\). What is the value of the product \(x \cdot y\)?

A -6

B -4

C 2

D 6

Texas Success Initiative Assessment 2.0 • Mathematics Practice Exam Page 3 of 6

Section 2: Algebraic Reasoning (Cont.) Questions 9 - 12

Q9

Which of the following is the equation of the line that is perpendicular to \(2x - 5y = 10\) and passes through the point \((4, -1)\)?

A \(y = -\frac{5}{2}x + 9\)

B \(y = -\frac{5}{2}x - 1\)

C \(y = \frac{5}{2}x - 11\)

D \(y = \frac{2}{5}x - 3\)

Q10

What are the solutions to the quadratic equation \(3x^2 - 10x - 8 = 0\)?

A \(x = -4 \text{ and } x = \frac{2}{3}\)

B \(x = 4 \text{ and } x = -\frac{2}{3}\)

C \(x = 2 \text{ and } x = -\frac{4}{3}\)

D \(x = -2 \text{ and } x = \frac{4}{3}\)

Q11

Simplify the expression \(\frac{(4x^3y^{-2})^2}{2x^{-4}y^3}\) and write it with only positive exponents.

A \(\frac{8x^{10}}{y^7}\)

B \(\frac{8x^2}{y^7}\)

C \(\frac{16x^{10}}{y^7}\)

D \(\frac{8x^{10}}{y}\)

Q12

A community garden is designed as a rectangle 12 meters wide by 20 meters long, with a semi-circular flower bed attached along one of the 12-meter widths. What is the total area of the garden to the nearest square meter?

A 257 m²

B 297 m²

C 313 m²

D 341 m²

Texas Success Initiative Assessment 2.0 • Mathematics Practice Exam Page 4 of 6

Section 3: Geometric & Spatial Reasoning Questions 13 - 16

Q13

A running track consists of a rectangular middle section with dimensions 100 meters by 50 meters and semi-circular ends on both of the 50-meter sides. What is the total perimeter (distance around the outside) of the track to the nearest meter?

A 357 meters

B 414 meters

C 514 meters

D 614 meters

Q14

A cylindrical water silo has a radius of 4 feet and a height of 15 feet. A cone-shaped roof with the same radius and a height of 6 feet is placed on top. What is the total volume of the combined structure, in terms of \(\pi\)?

A \(272\pi \text{ ft}^3\)

B \(288\pi \text{ ft}^3\)

C \(304\pi \text{ ft}^3\)

D \(336\pi \text{ ft}^3\)

Q15

A 15-foot ladder is leaning against a vertical wall. If the ladder makes a \(60^\circ\) angle with the flat ground, how high up the wall does the ladder reach, to the nearest tenth of a foot?

A 7.5 feet

B 10.6 feet

C 13.0 feet

D 26.0 feet

Q16

In a right-angled triangle \(ABC\), angle \(C\) is the right angle. If \(\cos A = \frac{5}{13}\) and side \(AC = 15\) units, what is the length of side \(BC\)?

A 12

B 36

C 39

D 45

Texas Success Initiative Assessment 2.0 • Mathematics Practice Exam Page 5 of 6

Section 4: Probabilistic & Statistical Reasoning Questions 17 - 20

Q17

Marcus has scores of 82, 85, 91, and 88 on his first four math exams. What score does he need to achieve on his fifth exam to earn a mean score of exactly 88 for all five exams?

A 88

B 92

C 94

D 96

Q18

The table below shows a survey of 200 high school students regarding preferred extracurricular activities. If a student is chosen at random, what is the probability they prefer Art, given they are an 11th grader?

A \(\frac{1}{8}\)

B \(\frac{1}{3}\)

C \(\frac{5}{11}\)

D \(\frac{25}{200}\)

Q19

A box contains 5 red, 4 blue, and 3 green marbles. If two marbles are drawn at random one after another without replacement, what is the probability that both marbles drawn are red?

A \(\frac{5}{33}\)

B \(\frac{25}{144}\)

C \(\frac{5}{12}\)

D \(\frac{5}{11}\)

Q20

A dataset consists of the values: 12, 15, 15, 18, 22, 25, 29, and 36. If a new value of 45 is added to this dataset, which of the following statements is true?

A The median increases by more than 2.

B The mean increases and the range increases by 9.

C The mode changes from 15 to 22.

D The mean increases and the range increases by 16.

Texas Success Initiative Assessment 2.0 • Mathematics Practice Exam Page 6 of 6

Question 7: Slope-Intercept Line Correct Option: A (\(y = -3x - 1\))

Explanation: First, calculate slope \(m\) from the given points \((-2, 5)\) and \((4, -13)\): \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-13 - 5}{4 - (-2)} = \frac{-18}{6} = -3 \] Substitute \(m = -3\) and point \((-2, 5)\) into point-slope form: \[ y - 5 = -3(x - (-2)) \implies y - 5 = -3x - 6 \implies y = -3x - 1 \]

Question 8: System of Equations Correct Option: A (-6)

Explanation: Use the elimination method:
Multiply \(3x - 4y = 18\) by \(3\) and \(2x + 3y = -5\) by \(4\): \[ \begin{cases} 9x - 12y = 54 \\ 8x + 12y = -20 \end{cases} \implies 17x = 34 \implies x = 2 \] Substitute \(x = 2\) back into second equation: \(2(2) + 3y = -5 \implies 3y = -9 \implies y = -3\).
Calculate product: \(x \cdot y = 2 \cdot (-3) = -6\).

TSIA2 Math Placement Prep • Solutions Guide Page 2 of 5

TSIA2 Math Solutions Guide

DETAILED ANSWER KEY

Questions 9 - 12

Question 9: Perpendicular Slopes Correct Option: A (\(y = -\frac{5}{2}x + 9\))

Explanation: Isolate \(y\) in \(2x - 5y = 10 \implies y = \frac{2}{5}x - 2\). The slope is \(m = \frac{2}{5}\). The perpendicular line slope is the negative reciprocal: \(m_{\perp} = -\frac{5}{2}\).
Substitute slope and \((4, -1)\) into point-slope form: \[ y - (-1) = -\frac{5}{2}(x - 4) \implies y + 1 = -\frac{5}{2}x + 10 \implies y = -\frac{5}{2}x + 9 \]

Question 10: Factoring Quadratics Correct Option: B (\(x = 4, x = -\frac{2}{3}\))

Explanation: Solve \(3x^2 - 10x - 8 = 0\) by factoring by grouping:
Find factors of \(3 \times (-8) = -24\) that add up to \(-10\) (these are \(-12\) and \(2\)): \[ 3x^2 - 12x + 2x - 8 = 0 \implies 3x(x - 4) + 2(x - 4) = 0 \implies (3x + 2)(x - 4) = 0 \] Set factors to zero: \(3x + 2 = 0 \implies x = -\frac{2}{3}\), and \(x - 4 = 0 \implies x = 4\).

Question 11: Exponential Simplification Correct Option: A (\(\frac{8x^{10}}{y^7}\))

Explanation: Apply power of a product rule first, then divide variable bases:
Numerator: \((4x^3y^{-2})^2 = 16x^6y^{-4}\).
Division: \[ \frac{16x^6y^{-4}}{2x^{-4}y^3} = 8 \cdot x^{6 - (-4)} \cdot y^{-4 - 3} = 8 \cdot x^{10} \cdot y^{-7} = \frac{8x^{10}}{y^7} \]

Question 12: Composite Area Correct Option: B (297 m²)

Explanation: Calculate individual areas and sum them up:
1. Rectangle: \(\text{Area} = 12 \times 20 = 240\text{ m}^2\).
2. Semi-circle: Diameter is \(12\text{ m}\), so radius \(r = 6\text{ m}\). \[ \text{Area}_{\text{semi}} = \frac{1}{2}\pi r^2 = \frac{1}{2}\pi (6^2) = 18\pi \approx 56.55\text{ m}^2 \] Total Area = \(240 + 56.55 = 296.55 \approx 297\text{ m}^2\) (to the nearest whole number).

TSIA2 Math Placement Prep • Solutions Guide Page 3 of 5

TSIA2 Math Solutions Guide

DETAILED ANSWER KEY

Questions 13 - 16

Question 13: Track Perimeter Correct Option: A (357 meters)

Explanation: The track boundary consists of two straight paths (top/bottom) and two semi-circles:
1. Straight paths: \(100\text{ m} + 100\text{ m} = 200\text{ m}\).
2. Combined semi-circular ends equal one full circle with diameter \(d = 50\text{ m}\): \[ C = \pi d = 50\pi \approx 157.08\text{ m} \] Total Perimeter = \(200 + 157.08 = 357.08 \approx 357\text{ meters}\).

Question 14: Combined Volumes Correct Option: A (\(272\pi \text{ ft}^3\))

Explanation: Find individual 3D solid volumes:
1. Cylinder: \(V_{\text{cylinder}} = \pi r^2 h = \pi (4^2)(15) = 240\pi\text{ ft}^3\).
2. Cone: \(V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (4^2)(6) = 32\pi\text{ ft}^3\). \[ V_{\text{total}} = 240\pi + 32\pi = 272\pi\text{ ft}^3 \]

Question 15: Trigonometry Height Correct Option: C (13.0 feet)

Explanation: Set up a right triangle where ladder represents hypotenuse (\(15\text{ ft}\)) and ground angle is \(60^\circ\). The wall height represents the opposite side: \[ \sin(60^\circ) = \frac{\text{Opposite}}{\text{Hypotenuse}} \implies \text{height} = 15 \times \sin(60^\circ) \] \[ \text{height} = 15 \times \frac{\sqrt{3}}{2} \approx 15 \times 0.866 \approx 12.99 \approx 13.0\text{ feet} \]

Question 16: Scaled Triangles Correct Option: B (36)

Explanation: Use the cosine ratio: \(\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AC}{AB} = \frac{5}{13}\). Given \(AC = 15\): \[ \frac{15}{AB} = \frac{5}{13} \implies AB = 39 \] Find the opposite side \(BC\) using Pythagorean theorem: \[ BC = \sqrt{39^2 - 15^2} = \sqrt{1521 - 225} = \sqrt{1296} = 36 \] Note: This is a 5-12-13 right triangle scaled up by a factor of 3 (\(15-36-39\)).

TSIA2 Math Placement Prep • Solutions Guide Page 4 of 5

TSIA2 Math Solutions Guide

DETAILED ANSWER KEY

Questions 17 - 20

Question 17: Mean with Missing Score Correct Option: C (94)

Explanation: Set up an equation for the mean of the five exam scores: \[ \frac{82 + 85 + 91 + 88 + x}{5} = 88 \] Simplify numerator and multiply both sides by 5: \[ 346 + x = 440 \implies x = 440 - 346 = 94 \] Marcus must score a 94 on his fifth exam.

Question 18: Conditional Probability Correct Option: B (\(\frac{1}{3}\))

Explanation: Calculate conditional probability \(P(\text{Art} \mid \text{11th Grader})\). Restrict the denominator sample space only to the 11th Grader row.
Total 11th Grade students = 75.
11th Grade students who prefer Art = 25. \[ P = \frac{25}{75} = \frac{1}{3} \]

Question 19: Dependent Events Correct Option: A (\(\frac{5}{33}\))

Explanation: Total marbles = \(5 \text{ red} + 4 \text{ blue} + 3 \text{ green} = 12\).
Probability of first red marble: \(P(R_1) = \frac{5}{12}\).
Since there is no replacement, 4 red and 11 total marbles remain.
Probability of second red marble: \(P(R_2 \mid R_1) = \frac{4}{11}\).
Combined probability: \[ P(R_1 \text{ and } R_2) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33} \]

Question 20: Statistical Range & Mean Correct Option: B (Mean increases, range increases by 9)

Explanation: Evaluate the impact of adding 45 to the original dataset:
Original range: \(\text{Max} - \text{Min} = 36 - 12 = 24\).
New range: \(45 - 12 = 33\) (Range increases by \(33 - 24 = 9\)).
Since 45 is significantly larger than the original mean of the numbers (which was 21.5), the overall mean will increase.

TSIA2 Math Placement Prep • Solutions Guide Page 5 of 5